Graphing of second derivatives.

[-Dragoon-]

Homework Statement

Given the following graph of h(x), identify:
1. The intervals where h(x) is increasing and decreasing.
2. The local maximum and minimum points of h(x)
3. The intervals where h(x) is concave up and concave down
4. The inflection point
5. Sketch the graph of h'(x) and h''(x).

Homework Equations

N/A

The Attempt at a Solution

I know I drew the graph really bad, but from my book the slope is 0 at x = 2. So the intervals of increase would be when x < 2 and x > 2, but what about the decrease since there is no negative slope at all? How would I determine the local maximums and local minimums in this scenario seeing as there is no decrease to a negative slope? What would be the inflection point in this case, if there aren't any local maximums and local minimums? Normally, I wouldn't have trouble drawing the derivative of this graph. However, it is a 3rd degree polynomial which would mean the derivative would be a parabola. How could I draw a parabola with only one value for x?

Any help? Thanks in advance.

 

[lineintegral1]

First of all, how do you know that the function has to be decreasing anywhere? Must it be increasing and decreasing in some interval defined on the domain? Or can it simply be increasing for all x? And does such a scenario constitute a local maximum at all?

Recall also that the inflection point is simply where the concavity changes. More formally, this is where the second derivative is equal to zero. Graphically, this can be found by plotting both of the derivatives (and analyzing the second). However, the change in concavity is fairly clear if you look at the original function. The inflection point does not necessarily depend on extrema.

How do you draw the derivative? Well, start figuring out the approximate derivative values for some values of x and plot them. What might help you get started is knowing that the derivative is zero at one point (which one?). Also, the parabola have y values that are one sign (positive or negative?).

 

[-Dragoon-]

Based on previous examples I've done, if a function has a local maximum or minimum it must have a negative slope at a point, correct? Therefore, in this case, it does not have a local maximum or minimum. Right?

For the intervals of increase and decrease, that would be x < 2 for decrease and x > 2 increase. Since there is only one point where the slope is zero (x = 2), this would mean that the parabola "bounces" off of the x-axis at that point but still as intervals of increase and decrease. Correct?

Now the intervals where concave up and concave down. I know the interval of concave up is x > 2, as this is where the it changes positive to negative. But what about the concave down? Would it be 0 < x < 2 or just x < 2? This is the only point that confuses me.

I was able to draw the graphs, so that was taken care of. Thanks for the reply, by the way.

 

[-Dragoon-]

Any help?

 

[lineintegral1]

Based on previous examples I've done, if a function has a local maximum or minimum it must have a negative slope at a point, correct? Therefore, in this case, it does not have a local maximum or minimum. Right?

For the intervals of increase and decrease, that would be x < 2 for decrease and x > 2 increase. Since there is only one point where the slope is zero (x = 2), this would mean that the parabola "bounces" off of the x-axis at that point but still as intervals of increase and decrease. Correct?

Now the intervals where concave up and concave down. I know the interval of concave up is x > 2, as this is where the it changes positive to negative. But what about the concave down? Would it be 0 < x < 2 or just x < 2? This is the only point that confuses me.

I was able to draw the graphs, so that was taken care of. Thanks for the reply, by the way.

Correct; it does not have a local max or min. But the function is increasing for all of x because the derivative is greater than or equal to zero for all of x. The derivative, however, is decreasing when x < 2. Yes, it hits the x-axis at 2 because the derivative is zero at that point.

Let's see if I can simplify concavity for you. If the geometry is confusing, simply take the second derivative. Wherever it is positive is where the original function is concave up. Likewise, wherever it is negative is where it is concave down. When the second derivative is zero, an inflection point occurs. So, draw the second derivative (should be linear) and see what it tells you. You're on the right track!